One-dimensional Schrödinger equation
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1 Chapter 1 One-dimensional Schrödinger equation In this chapter we will start from the harmonic oscillator to introduce a general numerical methodology to solve the one-dimensional, time-independent Schrödinger equation. The analytical solution of the harmonic oscillator will be first derived and described. A specific integration algorithm (Numerov) will be used. The extension of the numerical methodology to other, more general types of potentials does not present any special difficulty. For a particle of mass m under a potential V (x), the one-dimensional, timeindependent Schrödinger equation is given by: h2 d 2 ψ + V (x)ψ(x) = Eψ(x), (1.1) 2m dx2 where ψ(x) is the wave function, in general complex, and h is the Planck constant h divided by 2π. In the following we are focusing on the discrete spectrum: the set of isolated energy values for which Eq.(1.1) has normalizable solutions, localized in space. 1.1 The harmonic oscillator The harmonic oscillator is a fundamental problem in classical dynamics as well as in quantum mechanics. It represents the simplest model system in which attractive forces are present and is an important paradigm for all kinds of vibrational phenomena. For instance, the vibrations around equilibrium positions of a system of interacting particles may be described, via an appropriate coordinate transformation, in terms of independent harmonic oscillators known as normal vibrational modes. The same holds in quantum mechanics. The study of the quantum oscillator allows a deeper understanding of quantization and of its effects and of wave functions of bound states. In this chapter we will first remind the main results of the theory of the harmonic oscillator, then we will show how to set up a computer code that allows to numerically solve the Schrödinger equation for the harmonic oscillator. The resulting code can be easily modified and adapted to a different (not simply 5
2 quadratic) interaction potential. This will allow to study problems that, unlike the harmonic oscillator, do not have a simple analytical solution Units The Schrödinger equation for a one-dimensional harmonic oscillator is, in usual notations: d 2 ψ (E dx 2 = 2m h 2 1 ) 2 Kx2 ψ(x) (1.2) where K is the force constant (the force on the mass being F = Kx, proportional to the displacement x and directed towards the origin). Classically such an oscillator has a frequency (angular frequency) K ω = m. (1.3) It is convenient to work in adimensional units. These are the units that will be used by the codes presented at the end of this chapter. Let us introduce adimensional variables ξ, defined as ξ = ( ) mk 1/4 h 2 x = (using Eq.(1.3) for ω), and ɛ, defined as ( ) mω 1/2 x (1.4) h ε = Ē hω. (1.5) By inserting these variables into the Schrödinger equation, we find d 2 ( ) ψ dξ 2 = 2 ε ξ2 ψ(ξ) (1.6) 2 which is written in adimensional units Exact solution One can easily verify that for large ξ (such that ε can be neglected) the solutions of Eq.(1.6) must have an asymptotic behavior like ψ(ξ) ξ n e ±ξ2 /2 (1.7) where n is any finite value. The + sign in the exponent must however be discarded: it would give raise to diverging, non-physical solutions (in which the particle would tend to leave the ξ = 0 point, instead of being attracted towards it by the elastic force). It is thus convenient to extract the asymptotic behavior and assume ψ(ξ) = H(ξ)e ξ2 /2 (1.8) where H(ξ) is a well-behaved function for large ξ (i.e. the asymptotic behavior is determined by the second factor e ξ2 /2 ). In particular, H(ξ) must not grow like e ξ2, or else we fall back into a undesirable non-physical solution. 6
3 Under the assumption of Eq.(1.8), Eq.(1.6) becomes an equation for H(ξ): H (ξ) 2ξH (ξ) + (2ε 1)H(ξ) = 0. (1.9) It is immediate to notice that ε 0 = 1/2, H 0 (ξ) = 1 is the simplest solution. This is the ground state, i.e. the lowest-energy solution, as will soon be clear. In order to find all solutions, we expand H(ξ) into a series (in principle an infinite one): H(ξ) = A n ξ n, (1.10) n=0 we derive the series to find H and H, plug the results into Eq.(1.9) and regroup terms with the same power of ξ. We find an equation [(n + 2)(n + 1)A n+2 + (2ε 2n 1)A n ] ξ n = 0 (1.11) n=0 that can be satisfied for any value of ξ only if the coefficients of all the orders are zero: (n + 2)(n + 1)A n+2 + (2ε 2n 1)A n = 0. (1.12) Thus, once A 0 and A 1 are given, Eq.(1.12) allows to determine by recursion the solution under the form of a power series. Let us assume that the series contain an infinite number of terms. For large n, the coefficient of the series behave like A n+2 A n 2 n, that is: A n+2 1 (n/2)!. (1.13) Remembering that exp(ξ 2 ) = n ξ2n /n!, whose coefficient also behave as in Eq.(1.13), we see that recursion relation Eq.(1.12) between coefficients produces a function H(ξ) that grows like exp(ξ 2 ), that is, produces nonphysical diverging solutions. The only way to prevent this from happening is to have in Eq.(1.12) all coefficients beyond a given n vanish, so that the infinite series reduces to a finite-degree polynomial. This happens if and only if ε = n (1.14) where n is a non-negative integer. Allowed energies for the harmonic oscillator are thus quantized: ( E n = n + 1 ) hω n = 0, 1, 2,... (1.15) 2 The corresponding polynomials H n (ξ) are known as Hermite polynomials. H n (ξ) is of degree n in ξ, has n nodes, is even [H n ( ξ) = H n (ξ)] for even n, odd [H n ( ξ) = H n (ξ)] for odd n. Since e ξ2 /2 is node-less and even, the complete wave function corresponding to the energy E n : ψ n (ξ) = H n (ξ)e ξ2 /2 (1.16) 7
4 Figure 1.1: Wave functions and probability density for the quantum harmonic oscillator. has n nodes and the same parity as n. The fact that all solutions of the Schrödinger equation are either odd or even functions is a consequence of the symmetry of the potential: V ( x) = V (x). The lowest-order Hermite polynomials are H 0 (ξ) = 1, H 1 (ξ) = 2ξ, H 2 (ξ) = 4ξ 2 2, H 3 (ξ) = 8ξ 3 12ξ. (1.17) A graph of the corresponding wave functions and probability density is shown in fig Comparison with classical probability density The probability density for wave function ψ n (x) of the harmonic oscillator has in general n+1 peaks, whose height increases while approaching the corresponding classical inversion points (i.e. points where V (x) = E). These probability density can be compared to that of the classical harmonic oscillator, in which the mass moves according to x(t) = x 0 sin(ωt). The probability ρ(x)dx to find the mass between x and x + dx is proportional to the time needed to cross such a region, i.e. it is inversely proportional to the speed as a function of x: ρ(x)dx Since v(t) = x 0 ω cos(ωt) = ω x 2 0 x2 0 sin2 (ωt), we have ρ(x) dx v(x). (1.18) 1 x 2 0 x2. (1.19) This probability density has a minimum for x = 0, diverges at inversion points, is zero beyond inversion points. The quantum probability density for the ground state is completely different: has a maximum for x = 0, decreases for increasing x. At the classical inversion 8
5 point its value is still 60% of the maximum value: the particle has a high probability to be in the classically forbidden region (for which V (x) > E). In the limit of large quantum numbers (i.e. large values of the index n), the quantum density tends however to look similar to the quantum one, but it still displays the oscillatory behavior in the allowed region, typical for quantum systems. 1.2 Quantum mechanics and numerical codes: some observations Quantization A first aspect to be considered in the numerical solution of quantum problems is the presence of quantization of energy levels for bound states, such as for instance Eq.(1.15) for the harmonic oscillator. The acceptable energy values E n are not in general known a priori. Thus in the Schrödinger equation (1.1) the unknown is not just ψ(x) but also E. For each allowed energy level, or eigenvalue, E n, there will be a corresponding wave function, or eigenfunction, ψ n (x). What happens if we try to solve the Schrödinger equation for an energy E that does not correspond to an eigenvalue? In fact, a solution exists for any value of E. We have however seen while studying the harmonic oscillator that the quantization of energy originates from boundary conditions, requiring no nonphysical divergence of the wave function in the forbidden regions. Thus, if E is not an eigenvalue, we will observe a divergence of ψ(x). Numerical codes searching for allowed energies must be able to recognize when the energy is not correct and search for a better energy, until it coincides within numerical or predetermined accuracy with an eigenvalue. The first code presented at the end of this chapter implements such a strategy A pitfall: pathological asymptotic behavior An important aspect of quantum mechanics is the existence of negative kinetic energies: i.e., the wave function can be non zero (and thus the probability to find a particle can be finite) in regions for which V (x) > E, forbidden according to classical mechanics. Based on (1.1) and assuming the simple case in which V is (or can be considered) constant, this means d 2 ψ dx 2 = k2 ψ(x) (1.20) where k 2 is a positive quantity. This in turns implies an exponential behavior, with both ψ(x) exp(kx) and ψ(x) exp( kx) satisfying (1.20). As a rule only one of these two possibilities has a physical meaning: the one that gives raise to a wave function that decreases exponentially at large x. It is very easy to distinguish between the good and the bad solution for a human. Numerical codes however are less good for such task: by their very nature, they accept both solutions, as long as they fulfill the equations. If even 9
6 a tiny amount of the bad solution (due to numerical noise, for instance) is present, the integration algorithm will inexorably make it grow in the classically forbidden region. As the integration goes on, the bad solution will sooner or later dominate the good one and eventually produce crazy numbers (or crazy NaN s: Not a Number). Thus a nice-looking wave function in the classically allowed region, smoothly decaying in the classically forbidden region, may suddenly start to diverge beyond some limit, unless some wise strategy is employed to prevent it. The second code presented at the end of this chapter implements such a strategy. 1.3 Numerov s method Let us consider now the numerical solution of the (time-independent) Schrödinger equation in one dimension. The basic assumption is that the equation can be discretized, i.e. written on a suitable finite grid of points, and integrated, i.e. solved, the solution being also given on the grid of points. There are many big thick books on this subject, describing old and new methods, from the very simple to the very sophisticated, for all kinds of differential equations and all kinds of discretization and integration algorithms. In the following, we will consider Numerov s method (named after Russian astronomer Boris Vasilyevich Numerov) as an example of a simple yet powerful and accurate algorithm. Numerov s method is useful to integrate second-order differential equations of the general form d 2 y = g(x)y(x) + s(x) (1.21) dx2 where g(x) and s(x) are known functions. Initial conditions for second-order differential equations are typically given as y(x 0 ) = y 0, y (x 0 ) = y 0. (1.22) The Schrödinger equation (1.1) has this form, with g(x) 2m [E V (x)] and h 2 s(x) = 0. We will see in the next chapter that also the radial Schrödinger equations in three dimensions for systems having spherical symmetry belongs to such class. Another important equation falling into this category is Poisson s equation of electromagnetism, d 2 φ = 4πρ(x) (1.23) dx2 where ρ(x) is the charge density. In this case g(x) = 0 and s(x) = 4πρ(x). Let us consider a finite box containing the system: for instance, x max x x max, with x max large enough for our solutions to decay to negligibly small values. Let us divide our finite box into N small intervals of equal size, x wide. We call x i the points of the grid so obtained, y i = y(x i ) the values of the unknown function y(x) on grid points. In the same way we indicate by g i and s i the values of the (known) functions g(x) and s(x) in the same grid points. In order to obtain a discretized version of the differential equation (i.e. to obtain 10
7 an equation involving finite differences), we expand y(x) into a Taylor series around a point x n, up to fifth order: y n 1 = y n+1 = y n y n x y n( x) y n ( x) O[( x) 6 ] y n + y n x y n( x) y n ( x) O[( x) 6 ]. If we sum the two equations, we obtain: 24 y 24 y n ( x) 4 1 n ( x) y 120 y n ( x) 5 n ( x) 5 (1.24) y n+1 + y n 1 = 2y n + y n( x) y n ( x) 4 + O[( x) 6 ]. (1.25) Eq.(1.21) tells us that y n = g n y n + s n z n. (1.26) The quantity z n above is introduced to simplify the notations. The following relation holds: z n+1 + z n 1 = 2z n + z n( x) 2 + O[( x) 4 ] (1.27) (this is the simple formula for discretized second derivative, that can be obtained in a straightforward way by Taylor expansion up to third order) and thus y n z n = z n+1 + z n 1 2z n ( x) 2 + O[( x) 2 ]. (1.28) By inserting back these results into Eq.(1.25) one finds y n+1 = 2y n y n 1 + ( g n y n + s n )( x) ( g n+1y n+1 + s n+1 g n 1 y n 1 + s n 1 + 2g n y n 2s n )( x) 2 +O[( x) 6 ] (1.29) and finally the Numerov s formula [ ] [ ] ] ( x) y n g 2 ( x) n+1 = 2y n 1 5g 2 ( x) n y n 1 [1 + g 2 n (s n s n + s n 1 ) ( x) O[( x) 6 ] 12 (1.30) that allows to obtain y n+1 starting from y n and y n 1, and recursively the function in the entire box, as long as the value of the function is known in the first two points (note the difference with traditional initial conditions, Eq.(1.22), in which the value at one point and the derivative in the same point is specified). It is of course possible to integrate both in the direction of positive x and in the direction of negative x. In the presence of inversion symmetry, it will be sufficient to integrate in just one direction. In our case Schrödinger equation the s n terms are absent. It is convenient to introduce an auxiliary array f n, defined as ( x) 2 f n 1 + g n 12, where g n = 2m h 2 [E V (x n)]. (1.31) Within such assumption Numerov s formula can be written as y n+1 = (12 10f n)y n f n 1 y n 1 f n+1. (1.32) 11
8 1.3.1 Code: harmonic0 Code harmonic0.f90 1 (or harmonic0.c 2 ) solves the Schrödinger equation for the quantum harmonic oscillator, using the Numerov s algorithm above described for integration, and searching eigenvalues using the shooting method. The code uses the adimensional units introduced in (1.4). The shooting method is quite similar to the bisection procedure for the search of the zero of a function. The code looks for a solution ψ n (x) with a pre-determined number n of nodes, at an energy E equal to the mid-point of the energy range [E min, E max ], i.e. E = (E max + E min )/2. The energy range must contain the desired eigenvalue E n. The wave function is integrated starting from x = 0 in the direction of positive x; at the same time, the number of nodes (i.e. of changes of sign of the function) is counted. If the number of nodes is larger than n, E is too high; if the number of nodes is smaller than n, E is too low. We then choose the lower half-interval [E min, E max = E], or the upper half-interval [E min = E, E max ], respectively, select a new trial eigenvalue E in the mid-point of the new interval, iterate the procedure. When the energy interval is smaller than a pre-determined threshold, we assume that convergence has been reached. For negative x the function is constructed using symmetry, since ψ n ( x) = ( 1) n ψ n (x). This is of course possible only because V ( x) = V (x), otherwise integration would have been performed on the entire interval. The parity of the wave function determines the choice of the starting points for the recursion. For n odd, the two first points can be chosen as y 0 = 0 and an arbitrary finite value for y 1. For n even, y 0 is arbitrary and finite, y 1 is determined by Numerov s formula, Eq.(1.32), with f 1 = f 1 and y 1 = y 1 : The code prompts for some input data: y 1 = (12 10f 0)y 0 2f 1. (1.33) the limit x max for integration (typical values: 5 10); the number N of grid points (typical values range from hundreds to a few thousand); note that the grid point index actually runs from 0 to N, so that x = x max /N; the name of the file where output data is written; the required number n of nodes (the code will stop if you give a negative number). Finally the code prompts for a trial energy. You should answer 0 in order to search for an eigenvalue with n nodes. The code will start iterating on the energy, printing on standard output (i.e. at the terminal): iteration number, number of nodes found (on the positive x axis only), the current energy eigenvalue estimate. It is however possible to specify an energy (not necessarily an
9 eigenvalue) to force the code to perform an integration at fixed energy and see the resulting wave function. It is useful for testing purposes and to better understand how the eigenvalue search works (or doesn t work). Note that in this case the required number of nodes will not be honored; however the integration will be different for odd or even number of nodes, because the parity of n determines how the first two grid points are chosen. The output file contains five columns: respectively, x, ψ(x), ψ(x) 2, ρ cl (x) and V (x). ρ cl (x) is the classical probability density (normalized to 1) of the harmonic oscillator, given in Eq.(1.19). All these quantities can be plotted as a function of x using any plotting program, such as gnuplot, shortly described in the introduction. Note that the code will prompt for a new value of the number of nodes after each calculation of the wave function: answer -1 to stop the code. If you perform more than one calculation, the output file will contain the result for all of them in sequence. Also note that the wave function are written for the entire box, from x max to x max. It will become quickly evident that the code sort of works: the results look good in the region where the wave function is not vanishingly small, but invariably, the pathological behavior described in Sec.(1.2.2) sets up and wave functions diverge at large x. As a consequence, it is impossible to normalize the ψ(x). The code definitely needs to be improved. The proper way to deal with such difficulty is to find an inherently stable algorithm Code: harmonic1 Code harmonic1.f90 3 (or harmonic1.c 4 ) is the improved version of harmonic0 that does not suffer from the problem of divergence at large x. Two integrations are performed: a forward recursion, starting from x = 0, and a backward one, starting from x max. The eigenvalue is fixed by the condition that the two parts of the function match with continuous first derivative (as required for a physical wave function, if the potential is finite). The matching point is chosen in correspondence of the classical inversion point, x cl, i.e. where V (x cl ) = E. Such point depends upon the trial energy E. For a function defined on a finite grid, the matching point is defined with an accuracy that is limited by the interval between grid points. In practice, one finds the index icl of the first grid point x c = icl x such that V (x c ) > E; the classical inversion point will be located somewhere between x c x and x c. The outward integration is performed until grid point icl, yielding a function ψ L (x) defined in [0, x c ]; the number n of changes of sign is counted in the same way as in harmonic0. If n is not correct the energy is adjusted (lowered if n too high, raised if n too low) as in harmonic0. We note that it is not needed to look for changes of sign beyond x c : in fact we know a priori that in the classically forbidden region there cannot be any nodes (no oscillations, just decaying, or exploding, solutions). If the number of nodes is the expected one, the code starts to integrate inward from the rightmost points. Note the statement y(mesh) = dx: its only
10 goal is to force solutions to be positive, since the solution at the left of the matching point is also positive. The value dx is arbitrary: the solution is anyway rescaled in order to be continuous at the matching point. The code stops the same index icl corresponding to x c. We thus get a function ψ R (x) defined in [x c, x max ]. In general, the two parts of the wave function have different values in x c : ψ L (x c ) and ψ R (x c ). We first of all re-scale ψ R (x) by a factor ψ L (x c )/ψ R (x c ), so that the two functions match continuously in x c. Then, the whole function ψ(x) is renormalized in such a way that ψ(x) 2 dx = 1. Now comes the crucial point: the two parts of the function will have in general a discontinuity at the matching point ψ R (x c) ψ L (x c). This difference should be zero for a good solution, but it will not in practice, unless we are really close to the good energy E = E n. The sign of the difference allows us to understand whether E is too high or too low, and thus to apply again the bisection method to improve the estimate of the energy. In order to calculate the discontinuity with good accuracy, we write the Taylor expansions: yi 1 L = yl i y i L x y L i ( x) 2 + O[( x) 3 ] yi+1 R = yr i + y i R x y R i ( x) 2 + O[( x) 3 ] (1.34) For clarity, in the above equation i indicates the index icl. We sum the two Taylor expansions and obtain, noting that yi L = yi R = y i, and that y i L = y i R = y i = g iy i as guaranteed by Numerov s method: that is y L i 1 + y R i+1 = 2y i + (y R i y L i ) x g i y i ( x) 2 + O[( x) 3 ] (1.35) y i R y i L = yl i 1 + yr i+1 [2 g i( x) 2 ]y i + O[( x) 2 ] (1.36) x or else, by using the notations as in (1.31), y i R y i L = yl i 1 + yr i+1 (14 12f i)y i + O[( x) 2 ] (1.37) x In this way the code calculated the discontinuity of the first derivative. If the sign of y i R y i L is positive, the energy is too high (can you give an argument for this?) and thus we move to the lower half-interval; if negative, the energy is too low and we move to the upper half interval. As usual, convergence is declared when the size of the energy range has been reduced, by successive bisection, to less than a pre-determined tolerance threshold. During the procedure, the code prints on standard output a line for each iteration, containing: the iteration number; the number of nodes found (on the positive x axis only); if the number of nodes is the correct one, the discontinuity of the derivative y i R y i L (zero if number of nodes not yet correct); the current estimate for the energy eigenvalue. At the end, the code writes the final wave function (this time, normalized to 1!) to the output file. 14
11 1.3.3 Laboratory Here are a few hints for numerical experiments to be performed in the computer lab (or afterward), using both codes: Calculate and plot eigenfunctions for various values of n. It may be useful to plot, together with eigenfunctions or eigenfunctions squared, the classical probability density, contained in the fourth column of the output file. It will clearly show the classical inversion points. With gnuplot, e.g.: plot "filename" u 1:3 w l, "filename" u 1:4 w l (u = using, 1:3 = plot column3 vs column 1, w l = with lines; the second filename can be replaced by ). Look at the wave functions obtained by specifying an energy value not corresponding to an eigenvalue. Notice the difference between the results of harmonic0 and harmonic1 in this case. Look at what happens when the energy is close to but not exactly an eigenvalue. Again, compare the behavior of the two codes. Examine the effects of the parameters xmax, mesh. For a given x, how large can be the number of nodes? Verify how close you go to the exact results (notice that there is a convergence threshold on the energy in the code). What are the factors that affect the accuracy of the results? Possible code modifications and extensions: Modify the potential, keeping inversion symmetry. This will require very little changes to be done. You might for instance consider a double-well potential described by the form: [ (x ) 4 ( ) x 2 V (x) = ɛ 2 + 1], ɛ, δ > 0. (1.38) δ δ Modify the potential, breaking inversion symmetry. You might consider for instance the Morse potential: [ ] V (x) = D e 2ax 2e ax + 1, (1.39) widely used to model the potential energy of a diatomic molecule. Which changes are needed in order to adapt the algorithm to cover this case? 15
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